Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{n^2 + n - 72}{-5n - 45} \div \dfrac{-3n + 24}{-8n - 40} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{n^2 + n - 72}{-5n - 45} \times \dfrac{-8n - 40}{-3n + 24} $ First factor the quadratic. $p = \dfrac{(n - 8)(n + 9)}{-5n - 45} \times \dfrac{-8n - 40}{-3n + 24} $ Then factor out any other terms. $p = \dfrac{(n - 8)(n + 9)}{-5(n + 9)} \times \dfrac{-8(n + 5)}{-3(n - 8)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (n - 8)(n + 9) \times -8(n + 5) } { -5(n + 9) \times -3(n - 8) } $ $p = \dfrac{ -8(n - 8)(n + 9)(n + 5)}{ 15(n + 9)(n - 8)} $ Notice that $(n + 9)$ and $(n - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -8\cancel{(n - 8)}(n + 9)(n + 5)}{ 15(n + 9)\cancel{(n - 8)}} $ We are dividing by $n - 8$ , so $n - 8 \neq 0$ Therefore, $n \neq 8$ $p = \dfrac{ -8\cancel{(n - 8)}\cancel{(n + 9)}(n + 5)}{ 15\cancel{(n + 9)}\cancel{(n - 8)}} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $p = \dfrac{-8(n + 5)}{15} ; \space n \neq 8 ; \space n \neq -9 $